Graph y=-3x+7 on a graph

bill spent $19 at starbucks on a coffee and some cookies. if the coffee cost $5 each cookie cost $2, how many cookies did he buy

Nataliya [291]

Answer:

this is easy. we know that he spent $19 total, 1 coffee that costed $5, and that each cookie costs $2. all you have to do is subtract 5 from 19 and divide the remaining amount of money (14) by 2. this makes 7 the amount of cookies Bill bought.

6 0

3 years ago

to find the probability of flipping heads at least once if you flip a coin two times. The possible outcomes (we don't care about

nikitadnepr [17]

Answer:

The probability of flipping Heads at least once is $\frac{3}{4}$ .

Step-by-step explanation:

The probability of an event, say E, is the ratio of the favorable outcomes to the total number of outcomes, i.e.

$P (E) = \frac{Favorable\ outcomes}{Total\ outcomes}$

The sample space of flipping two coins is:

S = {HH, HT, TH and TH}

Total number of outcomes = 4

Compute the probability of flipping Heads at least once as follows:

Let X = heads.

P (X ≥ 1) = P (X = 1) + P (X = 2)

$=\frac{2}{4}+\frac{1}{4} \\=\frac{3}{4}$

Thus, the probability of flipping Heads at least once is $\frac{3}{4}$ .

The experiment of flipping a coin is a binomial experiment.

Since there are only two outcomes of the experiment, either a Heads or a Tails.

So if X is defined as the number of heads in n flips of a coin then the random variable X follows a binomial distribution with probability p = 0.5 of success.

6 0

3 years ago

A - P (1) = 3/8

crimeas [40]

A is true because the outcome of "1" happens three times (see the three tally marks) out of 8. So we have 3/8 as the probability for the outcome of "1".

B is false because the outcomes greater than "2" happen a total of 3 times out of 8. So it should be P(X > 2) = 3/8 and NOT P(X > 2) = 1/4

C is true as this is equivalent to what part A is saying. A is true so C is true. Being less than 2 is exactly identical as having an outcome of 1.

------------------------------------
In summary:

A = true
B = false
C = true

4 0

4 years ago

Assuming each area code is just as likely as any other, what's the probability that the area code of two back to back spam calls

sergij07 [2.7K]

Complete Question:

Yanay gets a lot of spam calls. An area code is defined to be a three digit number from 200-999 inclusive. In reality, many of these area codes are not in use, but for this question we'll simplify things and assume they all are.

Assuming each area code is just as likely as any other, what's the probability that the area code of two back to back spam calls are 804?

Answer:

1/6400 or 0.0000015625

Step-by-step explanation:

Given

Area Codes = 200 to 999 (inclusive)

Required

Determine the probability that back to back calls have area code 804

[First, we need to calculate the totality of area codes]

Total Area Codes = 999 - 200 + 1

Total Area Codes = 800

The probability of each code is 1/800

So, the probability of having back to back code of 804 is:

Probability = 1/800 * 1/800

Probability = (1/800)²

Probability = 1/640000

Hence, the required probability is 1/640000 or 0.0000015625

5 0

3 years ago

An electronic device factory is studying the length of life of the electronic components they produced. The manager selects two

Temka [501]

Answer:

The confidence interval will be given by:

$200000 \pm 44.72z$ , in which z is related to the confidence level.

For a confidence level of x%, z is the value in the z-table that has a pvalue of $1 - \frac{1 - z}{2}$

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In the context of this problems:

It means that the sampling distributions of the sample mean of 500 components will be approximated normal, with mean 200,000 and standard deviation $s = \frac{1000}{\sqrt{500}} = 44.72$

To build the confidence interval:

The confidence interval for the average length of life of the electronic components they produced will be given by:

$200000 \pm 44.72z$ , in which z is related to the confidence level.

For a confidence level of x%, z is the value in the z-table that has a pvalue of $1 - \frac{1 - z}{2}$

3 0

3 years ago